Question: $\dfrac{ -2f - 2g }{ -9 } = \dfrac{ -6f + 8h }{ 2 }$ Solve for $f$.
Multiply both sides by the left denominator. $\dfrac{ -2f - 2g }{ -{9} } = \dfrac{ -6f + 8h }{ 2 }$ $-{9} \cdot \dfrac{ -2f - 2g }{ -{9} } = -{9} \cdot \dfrac{ -6f + 8h }{ 2 }$ $-2f - 2g = -{9} \cdot \dfrac { -6f + 8h }{ 2 }$ Multiply both sides by the right denominator. $-2f - 2g = -9 \cdot \dfrac{ -6f + 8h }{ {2} }$ ${2} \cdot \left( -2f - 2g \right) = {2} \cdot -9 \cdot \dfrac{ -6f + 8h }{ {2} }$ ${2} \cdot \left( -2f - 2g \right) = -9 \cdot \left( -6f + 8h \right)$ Distribute both sides ${2} \cdot \left( -2f - 2g \right) = -{9} \cdot \left( -6f + 8h \right)$ $-{4}f - {4}g = {54}f - {72}h$ Combine $f$ terms on the left. $-{4f} - 4g = {54f} - 72h$ $-{58f} - 4g = -72h$ Move the $g$ term to the right. $-58f - {4g} = -72h$ $-58f = -72h + {4g}$ Isolate $f$ by dividing both sides by its coefficient. $-{58}f = -72h + 4g$ $f = \dfrac{ -72h + 4g }{ -{58} }$ All of these terms are divisible by $2$ Divide by the common factor and swap signs so the denominator isn't negative. $f = \dfrac{ {36}h - {2}g }{ {29} }$